Problem: Solve for $x$ and $y$ using substitution. ${-x+2y = 8}$ ${x = -2y-4}$
Since $x$ has already been solved for, substitute $-2y-4$ for $x$ in the first equation. ${-}{(-2y-4)}{+ 2y = 8}$ Simplify and solve for $y$ $2y+4 + 2y = 8$ $4y+4 = 8$ $4y+4{-4} = 8{-4}$ $4y = 4$ $\dfrac{4y}{{4}} = \dfrac{4}{{4}}$ ${y = 1}$ Now that you know ${y = 1}$ , plug it back into $\thinspace {x = -2y-4}\thinspace$ to find $x$ ${x = -2}{(1)}{ - 4}$ $x = -2 - 4$ ${x = -6}$ You can also plug ${y = 1}$ into $\thinspace {-x+2y = 8}\thinspace$ and get the same answer for $x$ : ${-x + 2}{(1)}{= 8}$ ${x = -6}$